PETE 411
Well Drilling
Lesson 26
* Well Control *
* Variable Geometry *
1
Well Control-Variable Geometry
 Initial Shut-In Conditions
 Driller’s Method
 Kick at Casing Seat
 Kick at Surface
 Wait and Weight Method
 Kick at Casing Seat
 Kick at Surface
2
Read:
Applied Drilling Engineering, Ch. 4
HW #14: Driller’s Method - due 11-11-02
HW #15: W&W Method - due 11-15-02
Quiz B
Thursday, Nov. 14, 7 - 9 p.m. Rm. 101
Closed Book
1 Equation sheet allowed, 8 1/2”x 11” (both sides)
3
Example Problem
1. Determine the pressure at the casing seat
at 4,000’ when using the old mud
(Driller’s) versus using the kill mud
(Wait and Weight) to circulate a gas
kick out of the hole.
2. Determine the casing pressure at the
surface when the top of the gas
bubble has just reached the surface,
for the same two mud weights used
above.
4
Example Problem
Well depth
Hole size
Drill pipe
= 10,000’
= 10.5”
= 4.5”, 16.60 #/ft
Drill Collars
= 8” * 3.5” * 500 ft
Surface casing = 4,000’, 13-3/8”, 68 #/ft
Mud Weight
= 10 ppg
Fracture gradient @ 4,000’ = 0.7 psi/ft
5
Example Problem
Shut in Annulus Pressure,
SICP = 400 psi
Shut in Drillpipe Pressure, SIDPP = 200 psi
Pit level increase = 20 barrels (kick size)
Drop the Z terms but consider Temperature
T at surface = 70 deg. F
Temperature gradient = 1.2 deg.F/100 ft
6
Initial (Closed-In) Conditions:
SIDPP  200 psi
SICP
 400 psi
Pit Level Increase
 20 bbl
Initial Mud Weight  10 # /gal
Initial mud gradient = 0.520 psi/ft
(0.052 * 10 = 0.520)
7
Initial (Closed-In) Conditions:
Bottom Hole Pressure,
P10,000
psi
 200  (10,000 ft ) * (0.520
)  5,400 psi
ft
Annular Vol/ft outside Drill Collars,
vDC_Ann

 gal  bbl 
2
2
2


 (10.5  8 ) in (12 in)
2 
4
 231in  42 gal 
VDC_Ann  0.04493 bbl/ft
( 20.28 ft/bbl )
8
vdp.csg = 0.13006 bbl/ft
4,000’
vdp,hole = 0.08743 bbl/ft
9,500’
vdc,hole = 0.04493 bbl/ft
10,000’
9
Height of Kick Fluid,
20 bbl
h10,000 
 445 ft
0.04493 bbl/ft
Hydrostatics in the Annulus,
5,400  400  0.520 * 9,555  Pkick_10,00 0
Hydrostatic Pressure across Kick Fluid,
Pkick_10,000  5,400  400  0.520 * 9,555
Pkick_10,000  31. 4 psi
10
Driller’s Method - kick at bottom
BHP  P0  Δ PK0  Δ PMA
- - - (2)
Weight of Kick Fluid  Pressure * area
lb 
2
2
2
 31.4 2 * (10.5  8 ) in
in 4
W  1,141 lb
F = P * A = W
11
SICP = 400 psi
4,000’
SIDPP = 200 psi
9,555’
9,500’
445’
10,000’
31.4
psi
PB = P 10,000 = 5,400 psi
12
Driller’s Method - kick at csg. seat
What is the pressure at 4,000 ft when the
top of the kick fluid first reaches that point?
V4,000  V10,000
 0.08743 * h4,000
 h4,000
 P10,000
* 
 P4,000
  T4,000
*
 T
  10,000




 5,400   70  48  460 
*
 20 * 

 
P
650

 4,000 
1,098,444

P4,000
13
Driller’s
Method
Top of
Kick at
Casing
Seat
4,000’
9,500’
10,000’
P = 5,400 psi
14
Driller’s Method - kick at csg. seat
Again,
BHP  P4,000  Δ PK_4,000  Δ PMA
PK_4,000
weight
1,141 lbs


π
area
10.5 2  4.5 2 in
4


P K_4,000 = 16.1 psi
15
Driller’s Method - kick at csg. seat
BHP  P4,000  Δ PK_4,000  Δ PMA
 5,400  P4,000  16  0.52 * (6,000  h4,000 )
5,384  P4,000  3,120  0.52 * h4,000
2,264  P4,000
 1,098,444 

 (0.52) * 

P
4,000


16
Driller’s Method - kick at csg. seat
This results in the quadratic Eqn:
P
2
4,000
 2,264 P4,000  571,191  0
With the solutions:
 P4,000


2,264  2,264  4 * 571,191

2
2
P4,000  2,493 psi  0.6233 psi/ft
0.5
 0.7
17
Driller’s Method - Top of Kick at Casing Seat
P0,ann = ?
P4,000 = 2,493 psi
4,000’
h4,000 = 441 ft
P = 16 psi
9,500’
10,000’
BHP = 5,400 psi

1,098,444 
h4,000 

P4,000 

18
Driller’s
Method
Top of
Kick at
Surface
We need two
equations…
P0 = ?
4,000’
9,500’
10,000’
BHP = 5,400 psi
19
Driller’s Method - kick at surface
When the bubble rises, it expands. The
volume of the bubble at the surface is given by:
 P10,000
V0  V10,000 
 P0
 T0

T
 10,000




(Z = const.)
 5400   70  460 
 
 (0.13006) h0  20 

 P0   70  120  460 
677,084
 h0 
Po
- - - (1)
20
Driller’s Method - kick at surface
Δ PK,0
weight
1,141 lb


π
area
2
2
2
12.415  4.5 in
4


Δ PK0  10.85  11 psi
21
Driller’s Method - kick at surface
From Eq. (2),
BHP  P0  Δ PK0  Δ PMA
5,400  P0  11  0.52 * (10,000  h0 )
But, from Eq. (1),
677,084
h0 
P0
22
Driller’s Method - kick at surface
BHP  P0  Δ PK0  Δ PMA
So,

677,084 

 5,400  P0  11  0.52 10,000 
P0


 5,400  5,200  11 P0  P0  352,084
2
P0  189 P0  352,084  0
2
Quadratic equation . . .
23
Driller’s Method - kick at surface


189  189  ( 4)(352,084)
 P0 
2
2
1
2
P0  695.34 psi  695 psi
677,084
 h0 
 973.74
695.34
h0  974 ft
4,000’
9,500’
10,000’
24
Driller’s Method - kick at surface
 P4,000  P0  0.52 * (4,000  974)  Δ PKO
 695  1,574  11
 2,280 psi
( = 0.57 psi/ft )
Alternativ ely,
4,000’
P4,000  P10,000  (0.52) * (10,000  4,000)
 5,400 - 3,120
 2,280 psi
9,500’
10,000’
25
Driller’s Method. Top of Kick at Surface
P0,ann = 695 psi
h0 = 974 ft
PK,0 = 11 psi
4,000’
P 4,000 = 2,280 psi
9,500’
10,000’
P10,000 = ?
26
Wait and Weight Method
- kick at casing seat
Calculate the pressure at 4,000 ft when the
top of the bubble reaches this point.
Volume of bubble at 4,000 ft
V4,000  V10,000
 P10,000

P
 4,000




 T4,000

T
 10,000




27
Wait and
Weight
Method
Top of
Kick at
Casing
Seat
Old Mud
Kill Mud
4,000’
9,500’
10,000’
BHP = 5,400 psi
28
W&W - Pressure at top of kick at 4,000 ft
V4,000  V10,000
 P10,000

P
 4,000




 T4,000

T
 10,000
 5,400 

(0.08743) h4,000  20 

P
 4,000 
1,098,444
 h4,000 
P4,000




 578 


 650 
- - - (6)
Also,
BHP  P4,000  ΔPK_4,000  ΔPM  ΔPM1
- - - (7)
29
Wait and Weight Method
- kick at surface
4,000’
Capacity inside drill string
= DP_cap. + DC_cap.
9,500’
10,000’
BHP = 5,400 psi
bbl
bbl




  0.01422
* 9,500 ft    0.0119
* 500 ft 
ft
ft




 141 bbl
#
 Quantity of 10.0
mud below the bubble.
gal
30
W&W - Pressure at top of kick
- kick at 4,000 ft
As before,
PK_4,000  16 psi
141 bbl
PM  0.052 * 10 *
0.08743 bbl/ft
PM  839 psi
SIDPP
Density of Kill Mud  Old Mud Wt. 
0.052 * 10,000
#
KWM  10.00  0.38  10.38
gal
31
W&W - Pressure at top of kick
- kick at 4,000 ft
141 

ΔPM1  0.052 * 10.38 *  6,000  h4,000 

0.08743 

BHP  P4,000  ΔPK_4,000  ΔPM  ΔPM1
5,400  P4,000  16  839  2,368  (0.5398) h4,000
32
W&W - Pressure at top of kick
- kick at 4,000 ft
 2,177  P4 000
 1,098,444 

 (0.5398)

P
4,000


2
 P4,000
 2,177 P4,000  592,940  0
 P4,000


2,177  2,177  4 * 592,940

2
P4,000  2,422 psi
2

0 ,5
 0.61 psi/ft
33
Wait and
Weight
Method
Top of
Kick at
Surface
4,000’
9,500’
Old Mud
Kill Mud
10,000’
34
Wait and Weight Method
- kick at surface
Volume of gas bubble at surface:
 P10,000
V0  V10,000 
 P0
 T0

T
 10,000




 5,400  530 

 0.13006 * h0  20

 P0  650 
677,084
 h0 
P0
35
Wait and Weight Method
(Engineer’s Method) - kick at surface
P10,000  P0  ΔPK0  ΔPM  ΔPM1
As before,
ΔPK,0  11 psi
Assume all 10 lb/gal mud is inside the
13 3/8” csg. Then the height of 10 lb mud:
141.0 bbl
hM 
 1,084 ft
0.13006 bbl/ft
36
Wait and Weight Method
(Engineer’s Method) - kick at surface
Hydrostatic head across the mud columns:
Old Mud:
 ΔPM  0.052 * 10 * 1,084  564 psi
Kill Mud:
PM1  0.052 * 10.38 * (10,000  h0  1,084)
37
Hydrostatics in Annulus
W&W Method - kick at surface
P10,000  P0  ΔPK0  ΔPM  ΔPM1
5,400  P0  11  564  0.5398 * (8,916  h0 )
12.14  P0  (0.5398) h0
38
Wait and Weight Method
kick at surface
From Eq. 4, substituting for h0
 677,084 

 12  P0  (0.5398) 
P0


 P02  12P0  365,490  0
 P0 

12  12 2  4 * 365,490

1
2
2
P0  610.59  611 psi
39
Wait and Weight Method
- kick at surface
Height of Bubble at Surface
677,084 677,084
h0 

P0
610.59
 1,109 ft
40
Check Pressure at 4,000 ft
- kick at surface
 P4,000  P0  ΔPK0  ΔPM  ΔPM1
 611  11  569 
 0.052 * 10.38 * (4,000 - 1,109 - 1,093)
 2,161 psi
 0.54 psi/ft
Looks OK
41
Wait and Weight Method
Top of Kick at Surface
Old Mud
P0,ann = 611 psi
Kill Mud
h0 = 1,109 ft
PK,0 = 11 psi
4,000’
9,500’
10,000’
POld Mud = 569 psi
P 4,000 = 2,161 psi
P10,000 = ?
42
Summary
Bubble at 10,000 ft
Driller’s
Method
P4,000
P0
Engineer’s
Method
2,480
2,480
400
400
43
Summary
Top of Bubble at 4,000 ft
Driller’s
Method
P4,000
P0
Engineer’s
Method
2,493
2,422
413
342
44
Summary
P4,000
P0
Top of Bubble at surface
Driller’s
Method
Engineer’s
Method
2,280
2,161
695
611
45
46
Why the difference?
47
Wait and Weight Method
Kick Intensity, ppg
48
Kick Intensity, ppg
Pump Strokes
49
Volume of Mud Pumped, bbl
50